Find the value of $\int_0^{\infty} \frac{x^{a-1}\,\mathrm{d}x}{1+x^{2}},\,$ where $\,0

Question

Using the Residue Theorem calculate $$ \int_0^{\infty} \frac{x^{a-1}\,\mathrm{d}x}{1+x^{2}},\,\,\,\, \text{where}\,\,\,0<a<2. $$ My solution comes out to be
$$ \frac{\pi}{2}[i^{a-1}-(-i)^{a-1}]$$

How to proceed from here since the solution of a rational expression must be independent of $i$

Write $i^{a-1}=e^{(a-1)\ln(i)}=e^{(a-1)\pi i/2}=\cos((a-1)\pi/2)+i\sin((a-1)\pi/2)$. Do the same with $(-i)^{a-1}$. Likely the $i$'s simplify. — OR., Jun 25 '17 at 12:07
$$ i^{a−1}=e^{(a−1)ln(i)}$$ how did this come to be? i'm sorry but i don't understand this — Irrational_Number, Jun 25 '17 at 12:16
It is the property of logarithm of being the inverse of the exponential. $e^{\ln(x)}=x$. Put $x=i^{a-1}$. You get $i^{a-1}=e^{\ln(i^{a-1})}$. Then use that $\ln(A^B)=B\ln(A)$. — OR., Jun 25 '17 at 12:29
using this method i'll just get the form of $e^{ix}-e^{-ix}$ which would be $2isinx$, the $i$ would still persist — Irrational_Number, Jun 25 '17 at 13:08
But at least one $i$ remains outside of the exponentials:$i^{a-1}=-ie^{i \pi /2 a}$ — N74, Jun 25 '17 at 13:33

Yiorgos S. Smyrlis · Accepted Answer · 2017-06-26T19:54:45.203

I imagine you have integrated $\,\,f(z)=\dfrac{z^{a-1}}{z^2+1},\,$ along the contour $\,\gamma=\gamma_1\cup\gamma_2\cup\gamma_3\cup\gamma_4,\,$ where $$ \gamma_1=[-R,-\varepsilon],\quad\gamma_2=\{\varepsilon\mathrm{e}^{i(\pi-t)}: t\in[0,\pi]\},\quad \gamma_3=[\varepsilon,R],\quad\gamma_4=\{R\mathrm{e}^{it}: t\in[0,\pi]\}. $$ In such case $$ z^{a-1}=\mathrm{e}^{(a-1)\log z}, $$ where $\log z$ is a branch of logarithm defined in $$ \Omega=\mathrm{C}\setminus \{it: t\in (-\infty,0]\}, $$ as follows. If $z=r\mathrm{e}^{i\vartheta}$, then $\log z=\log r+i\vartheta$, with $$ \vartheta\in (-\pi/2,3\pi,2). $$ So, in such case, Residue Theorem provides $$ \int_\gamma f(z)\,dz=2\pi i\,\mathrm{Res}\left(\frac{z^{a-1}}{1+z^2},i\right)=2\pi i\cdot\frac{i^{a-1}}{2i}=\pi\mathrm{e}^{i(a-1)\pi/2}=-i\pi \mathrm{e}^{ia\pi/2}. $$ Clearly $$ \lim_{R\to\infty,\,\varepsilon\to 0}\int_{\gamma_1}f(z)\,dz= \int_{-\infty}^0\frac{z^{a-1}}{z^2+1}=(-1)^{a-1}\int_0^\infty\frac{x^{a-1}\,dx}{1+x^2}=-\mathrm{e}^{ia\pi}\int_0^\infty\frac{x^{a-1}\,dx}{1+x^2}, $$ since $(-1)^{a-1}=\mathrm{e}^{i(a-1)\pi}=-\mathrm{e}^{ia\pi},\,$ and $$ \lim_{R\to\infty,\,\varepsilon\to 0}\int_{\gamma_3}f(z)\,dz= \int_0^{\infty}\frac{z^{a-1}}{z^2+1}=\int_0^\infty\frac{x^{a-1}\,dx}{1+x^2}. $$ Meanwhile, $\lim_{\varepsilon\to 0}\int_{\gamma_2}f(z)\,dz=0$ and $\lim_{R\to\infty}\int_{\gamma_4}f(z)\,dz=0$. Hence, altogether $$ (1-\mathrm{e}^{ia\pi})\int_{0}^\infty\frac{x^{a-1}\,dx}{1+x^2}= \lim_{\varepsilon\to 0,\,\,R\to\infty}\int_\gamma f(z)\,dz= 2\pi i\,\mathrm{Res}\left(\frac{z^{a-1}}{1+z^2},i\right) =-i\pi \mathrm{e}^{ia\pi/2}. $$ Hence $$ \int_{0}^\infty\frac{x^{a-1}\,dx}{1+x^2}=\frac{i\pi \mathrm{e}^{ia\pi/2}}{\mathrm{e}^{ia\pi}-1}=\frac{i\pi }{\mathrm{e}^{ia\pi/2}-\mathrm{e}^{-ia\pi/2}}=\frac{\pi}{2\sin (a\pi/2)}. $$

Thanks for the help – Irrational_Number Jun 26 '17 at 18:11 — Irrational_Number, Jun 26 '17 at 18:11

robjohn · Answer 2 · 2017-06-26T18:13:16.177

4

In this answer, it is shown that for $m\gt0$ and $-1<n<m-1$ $$ \frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x $$ Plug in $n=a-1$ and $m=2$ to get that for $0\lt a\lt2$ $$ \int_0^\infty\frac{x^{a-1}}{1+x^2}\,\mathrm{d}x=\frac\pi2\csc\left(\frac{\pi a}2\right) $$

edited Jun 26 '17 at 18:13

answered Jun 26 '17 at 17:00

robjohn

345,667

Is the downvote because I cited another residue theorem/contour integration answer? If so, I believe it is better simply to cite a previous answer than to replicate it everywhere it applies. – robjohn Jun 26 '17 at 18:03
+1. I agree with you. We don't have to "reinvent the wheel". – Felix Marin Jun 28 '17 at 02:03

Find the value of $\int_0^{\infty} \frac{x^{a-1}\,\mathrm{d}x}{1+x^{2}},\,$ where $\,0

2 Answers2